Number of ways to choose $4$ objects out of $6$ groups with $3$ members each

Question

Suppose a box contains 18 balls number 1-6, three balls with each number. When 4 balls are drawn without replacement, how many outcome are possible?

I took $6\choose4$ ways of picking the balls and then took away the possibilities where 4 of the same ball were chosen as this is impossible (can be done in 6 ways)

This gives me a final answer of; $${6\choose4} - 6$$

My question is; Is this method/answer correct and also is there a general way to solve this type of problem?

Note; This problem is from a book called 'An introduction to Combinatorics and Graph Theory' and I think it comes under the umbrella of multisets and upper bounds.

Answer 1

Let $x_k$ denote the number of balls numbered $k$ that are selected. If four of the $18$ balls are selected, then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4 \tag{1}$$ where $x_k \leq 3$ for $1 \leq k \leq 6$. Equation 1 is an equation in the non-negative integers. A particular selection corresponds to a choice of where to place five addition signs in a row of four ones. For instance, $$1 + 1 + + 1 + + 1$$ corresponds to $x_1 = 1$, $x_2 = 1$, $x_3 = 0$, $x_4 = 1$, $x_5 = 0$, and $x_6 = 1$. Hence, if there were no restrictions, the number of solutions of equation 1 would be the number of ways five addition signs could be inserted into a row of four ones, which is
$$\binom{4 + 5}{5} = \binom{9}{5}$$ since we must choose which five of the nine symbols (four ones and five addition signs) will be addition signs. However, six of these solutions violate the restriction that $x_k \leq 3$ for $1 \leq k \leq 6$, namely those in which $x_k = 4$ for some $k$ satisfying $1 \leq k \leq 6$. Hence, the number of ways to select four balls from $18$ balls consisting of three balls apiece numbered from 1 - 6 is $$\binom{9}{5} - 6$$

Answer 2

$\binom{6}{4}-6$ counts ways to select four colours minus ways to select one colour.

That's not right at all.

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You wish to count the distinct numbers and colours that may be selected.

As to numbers, you can select from restricted partitions of 4. Specifically: $(3:1), (2:2), (2:1:1), (1:1:1:1)$

That is you may select either: a triple-and-single, two-doubles, a double-and-two-singles, or four-singles.

But how many ways can you select colours for each of these alternatives?   (There's a lot more than nine)

An alternative approach is to use the stars-and-bars method to count ways to put four identical balls into six bins such that no more than three go into the same bin.

$$\mathop{\underline{\binom{6}{1,1,4}}}_\textrm{#triple-&-single}+\mathop{\underline{binom{6}{2,4}}}_\textrm{#two-doubles}+\mathop{\underline{\binom{6}{1,2,3}}}}_\textrm{#double-&-two-singles}+\mathop{\underline{\binom{6}{4,2}}}_\textrm{#four-singles} = \underline{\binom{4+6-1}{6-1}-6}$$